Module M5.5 Further integration

Transcription

1 F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S Module M5.5 Further integration Opening items. Module introduction. Fast track questions.3 Ready to study? Further techniques of integration. Partial fractions. Completing the square.3 Splitting the numerator.4 Substitutions involving hyperbolic functions.5 Using trigonometric identities.6 Reduction formulae.7 Mixed examples 3 Some definite integrals 3. Integrals with an infinite upper or lower limit 3. Gaussian integrals 3.3 Line integrals 4 Closing items 4. Module summary 4. Achievements 4.3 Exit test Exit module COPYRIGHT 998 S57 V.

2 Opening items. Module introduction This module discusses in depth a range of techniques which will enable you to evaluate a wide range of integrals. Such a detailed treatment may not be relevant to your course of study, and you are therefore advised to consult your tutor before working through the module. You should be prepared to spend more time than for the other FLAP modules if you are advised to read all the material. In Section we discuss several ingenious tricks which you can combine with methods such as integration by parts and integration by substitution in order to evaluate a very wide range of integrals. Some of these tricks (partial fractions, completing the square and splitting the numerator) involve algebraic manipulation of the integrand, while others make use of trigonometric identities to simplify integrals of powers of trigonometric functions. We also discuss some particularly useful substitutions involving hyperbolic functions. Subsection 3. deals with certain types of improper integral those with an infinite upper or lower limit. We will explain there how to define and evaluate these. Subsection 3. discusses some integrals of this sort known as Gaussian integrals, which arise very often in physics (in quantum mechanics and in the kinetic theory of gases, for example). We show how these can all be evaluated in terms of the basic Gaussian integral exp ( x ) dx. COPYRIGHT 998 S57 V.

3 Finally, in Subsection 3.3, we introduce and explain the idea of a line integral an integral of the form rb F(r) dr. Integrals of this sort arise in calculating the work done by a force, for example, or the electrostatic ra potential difference between two points in a region of space where there is an electric field. You may occasionally find that your answers to the exercises differ from ours. This may be because your answer is in a slightly different form; for example, you may have written log e x (x ) where we have written log x log (x ). If you cannot tell if your expression for an indefinite integral is the same as ours, you e e can always check your answer by differentiating it. Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the Fast track questions given in Subsection.. If not, proceed directly to Ready to study? in Subsection.3. COPYRIGHT 998 S57 V.

4 . Fast track questions Study comment Can you answer the following Fast track questions?. If you answer the questions successfully you need only glance through the module before looking at the Module summary (Subsection 4.) and the Achievements listed in Subsection 4.. If you are sure that you can meet each of these achievements, try the Exit test in Subsection 4.3. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module. Question F Find the integrals (a) x dx, (b) dx + 8x 4x 4x COPYRIGHT 998 S57 V.

5 Question F x+3 dx Evaluate the definite integral ( x + ) ( x + ) Question F3 Evaluate the definite integral exp ( 3x + 4x) dx 4given that4 exp ( y ) dy = COPYRIGHT 998 π. S57 V.

6 Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to Ready to study? in Subsection.3. Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Closing items. COPYRIGHT 998 S57 V.

7 .3 Ready to study? Study comment In order to study this module, you will need to be familiar with the following terms: completing the square, definite integral, even function, hyperbolic function, improper integral, integrand, integration by parts, integration by substitution, inverse hyperbolic function, limits of integration, scalar product and vector. If you are uncertain of any of these terms, you can review them now by referring to the Glossary which will indicate where in FLAP they are developed. In addition, you will x n + e ax need to be familiar with standard integrals (such as x n dx = + C and e ax dx = + C ), and know how to n + a evaluate definite and indefinite integrals by the method of substitution, or by integration by parts. You will also need to be familiar with trigonometric identities, and with the analogous identities involving hyperbolic functions. The following Ready to study questions will allow you to establish whether you need to review some of these topics before embarking on this module. COPYRIGHT 998 S57 V.

8 Question R + can be written as a single fraction by putting x x + x+ (x ) + = + x x + (x )(x + ) (x )(x + ) (x + ) + (x ) 3x = = (x )(x + ) (x )(x + ) The expression Use a similar method to express the following as single fractions: (a) x 5 5, (b) +, (c) (x 3) (x + 3) x 3 + x 9(x ) 9(x + ) 3(x + ) Question R Write the following quadratic functions in completed square form: (a) 3x x +6,4(b) 3 4x x COPYRIGHT 998 S57 V.

9 Question R3 (a) If y = cos(x), express sin4x in terms of y. (b) If y = cosh(x), express sinh4 x in terms of y. Question R4 Find the indefinite integrals: dx,4(c) (a) dx. dx,4(b) + 3x 4 + 9x 5 4x COPYRIGHT 998 S57 V.

10 Question R5 Evaluate the definite integrals: (a) π x sin (3x) dx,4(b) x + 4x dx. Question R6 Define the hyperbolic functions coshx and sinhx, and use your definitions to prove the identity coshx sinhx =. COPYRIGHT 998 S57 V.

11 Question R7 (a) What are the derivatives of coshx and sinhx? Use these derivatives, and the quotient rule, to find the derivative of tanhx. (b) Find the integrals cosh ( x) dx and sinh (x 3) dx. Question R8 If r is the vector xi + y j + zk, a is the vector i k and b is the vector i j + k, find an expression for (a b)(b r) in terms of x, y and z. COPYRIGHT 998 S57 V.

12 Further techniques of integration. Partial fractions dx. The integrand does not seem to be a particularly Suppose that we want to find the integral (x + )(x + ) complicated function of x. Yet this is not an integral that can immediately be related to a standard integral, nor will integration by parts work (as you will find if you try!), nor is it easy to find a substitution that will simplify the integral. However, this integral can in fact be found quite easily if we first split the integrand into its partial fractions. To do this, we write a b = + (x + )(x + ) x + x + () where a and b are constants that we will need to find. Adding together the two fractions on the right-hand side of Equation gives us COPYRIGHT 998 S57 V.

13 (a + b)x + a + b = (x + )(x + ) (x + )(x + ) from which we deduce that a + b = and a + b =, so that a =, b =. Thus we have the identity = (x + )(x + ) x + x + It follows that dx = dx dx = log e (x + ) log e (x + ) + C x + x+ (x + )(x + ) dx can be applied to a wide variety of integrals of The method we have used here to find (x + )(x + ) fractions whose denominators can be factorized. Here is a slightly more complicated example. COPYRIGHT 998 S57 V.

14 Example x dx. Find x +x 6 Solution4Since the denominator is not already factorized, we must first find its factors, which are (x ) and x x (x + 3). We now express = in partial fractions. We write x + x 6 (x )(x + 3) x a b (a + b)x + 3a b = + = (x )(x + 3) x x + 3 (x )(x + 3) from which we deduce that a + b = and 3a b =, i.e. a = /5 and b = 3/5. COPYRIGHT 998 S57 V.

15 Thus so that x 3 = + x + x 6 5(x ) 5(x + 3) 3 x 3 dx = log e (x ) + log e (x + 3) + C dx + dx = x + x x x This example shows that using partial fractions to evaluate integrals generally involves three steps: Step 4Factorize the denominator of the integrand (if necessary). Step 4Express the integrand in terms of partial fractions. Step 34Integrate each partial fraction. Practise these steps by doing the following question. Question T dx.4 Find the integral 9 4x COPYRIGHT 998 S57 V.

16 The three steps described above apply equally well if there is a repeated factor in the denominator of the integrand. For example, consider the integral dx 3 x x + x The denominator here is x 3 x + x = x(x x + ) = x(x ), which has a repeated factor (x ). We now write A term for ( x ) A term for ( x ) } 67 a b d ( a + b ) x + ( d a b ) x + a = + + = x x(x ) x (x ) x(x ) (and notice that we have included a term for (x ) and a term for (x ) ) so that a + b =, d a b =, and a =, i.e. a =, b =, d =. So = + 3 x x + x x x (x ) COPYRIGHT 998 S57 V. ()

17 dx. Find 3 x x + x Partial fractions can also often be used to integrate a fraction even if its denominator does not factorize completely into real linear factors, as in the following example. Example x+3 dx Find the integral (x )(x + ) Solution4We cannot factorize the quadratic factor (x + ). So, to express the integrand in terms of partial fractions, we must first write it in the form Two constants for the quadratic factor x+3 a = + (x )(x + ) x 678 bx + d x + = (a + b)x + (d b)x + a d (x )(x + ) (and notice that the term corresponding to the factor (x + ) includes two unknown constants) from which it follows that a + b =, d b = and a d = 3, i.e. a =, b =, d =. COPYRIGHT 998 S57 V.

18 x + x+3 dx = dx dx So x x + (x )(x + ) (3) The first integral on the right-hand side of Equation 3 is equal to loge(x ) + C. To evaluate the second x dx + dx. With the substitution u = x +, we integral, we split it into two, and write it as x + x + x dx = log e (x + ) + C, while dx is a standard integral, equal to arctanx + C. quickly find that x + x + Substituting these results into Equation 3 gives us x+3 dx = log e (x ) log e (x + ) arctan x + C 4 (x )(x + ) COPYRIGHT 998 S57 V.

19 Question T 3x + dx.4 Find the integral (x 3)( x + ) The technique of splitting the integrand into partial fractions will enable you to find many integrals of the form p(x) dx, where p(x) and q(x) are both polynomials in x. However, it clearly will not be of any use unless q(x) q(x) factorizes, at least partly. What can we do if q(x) does not factorize at all? This case is discussed in the next two subsections. COPYRIGHT 998 S57 V.

20 . Completing the square If you are asked to find the integral dx, your response may now be to try writing the integrand in x + 4x + 8 terms of partial fractions. If you try this, you will quickly find that the quadratic equation x + 4x + 8 = has no real roots, so that we cannot factorize the integrand, and partial fractions are of no use. To see how to proceed with an integral of this sort, recall that there are integrals similar to this which you do dx. This also has the property that its integrand does know how to evaluate; for example, the integral x +4 not factorize. However it can be quickly evaluated by means of the substitution x = tanu. Then x + 4 = 4 sec u, and dx = sec u du, so that the integral becomes du = u+c= arctan (x ) + C dx look very much like the integral dx by the trick known as We can make the integral x + 4x + 8 x +4 completing the square, in which we write the denominator as the sum of two squared terms, one involving x and the other a constant. In the case under consideration, it works as follows. COPYRIGHT 998 S57 V.

21 We notice that x + 4x = (x + ) 4, so that x + 4x + 8 = (x + ) + 4. dx Thus we have dx = x +4+8 (x + ) + 4 We now make the change of variables y = x +, so as to obtain dy dx = x + 4x + 8 y +4 We have just evaluated the integral on the right-hand side here; it is equal to Replacing y by x +, we finally find that dx = arctan[(x + ) ] + C x + 4x + 8 arctan (y ) + C. Here is a slightly more complicated example, which we will set out as a series of steps; you can follow these when you do similar problems. COPYRIGHT 998 S57 V.

22 Example 3 dx Find the integral I = x + 6x + 5 Solution4Step 4First, write the denominator in completed square form. [ ] x + 6x + 5 = (x + 3x) + 5 = ( x + 3 ) = ( x + 3 ) + dx So I = [x + (3 )] + 3 dy, then I = + ( ) y Step 34Make the substitution y = tan u, then y + = sec u and dy = sec u du. So I= du = du = u + C. Step 44Finally, express the integral in terms of x (using the fact that x + 3 = tan u ), I = u + C = arctan(y) + C = arctan(x + 3) + C.4 Step 4Make the substitution y = x + COPYRIGHT 998 S57 V.

23 Question T3 dx 4 Find the integral I = 3x 6x + 7 The technique of completing the square can also be used in conjunction with a substitution of the form x = asinu, where a is a constant; this enables us to find integrals such as I = dx. To evaluate 3 x x this integral, we first write the quadratic expression appearing under the square root sign in completed square form. Write 3 x x in completed square form. So dx. We now make the substitution y = x + ; then the integral dx = 3 x x 4 (x + ) dy. This integral can easily be found by the substitution y = sinu, so that dy = cosudu, becomes 4 y COPYRIGHT 998 S57 V.

24 then dy = 4 y ( cos u du) 4 ( sin u) ( cos u du) = du = u + C = arcsin (y ) + C = cos u Replacing y by x +, we finally have x + +C dx = arcsin 3 x x Here is another example, which again we will set out as a series of steps. COPYRIGHT 998 S57 V.

25 Example 4 Find the integral I = dx. 6x x Solution4Step 4Write 6x x in completed square form: 6x x = (x 6x) = [(x 3) 9] = 9 (x 3) so I= dx 9 (x 3) dy Step 4Make the substitution y = x 3; then I = 9 y Step 34Make the substitution y = 3sinu; then 9 y = 3cosu and dy = 3cosudu. So I = du = u + C Step 44Express I in terms of x: x 3 y + C 4 I = u + C = arcsin + C = arcsin 3 3 COPYRIGHT 998 S57 V.

26 Question T4 Find the integral I = dx x x In this subsection, we have so far discussed only indefinite integrals. However the techniques presented here can just as easily be used to evaluate definite integrals. You need only remember to transform the limits of integration appropriately in each of Steps and 3; if you do this, there will be no need for Step 4. Question T5 7 dx.4 Evaluate x 6x COPYRIGHT 998 S57 V.

27 .3 Splitting the numerator In the previous subsection, you learnt how to find integrals where the integrand had a quadratic denominator that did not factorize, and had a constant in the numerator. Suppose the integrand had an expression linear in x in the x + 3 numerator, instead of a constant; for example, suppose you were required to integrate. x 4x + How would you proceed here? To see what to do in such a case, notice that we in fact already encountered an integral of this sort in Example, x + dx. We evaluated it by writing it as the sum of the two integrals: where we had to find the integral x + x dx and dx. The first of these integrals could be easily evaluated, since the numerator is equal x + x + to the derivative of the denominator; thus the substitution y = x + enabled us to find it. The second could also be found, using the substitution x = tany. COPYRIGHT 998 S57 V.

28 x + 3 We can apply the same idea to an integral such as dx. We write the numerator as the sum of two x 4x + terms, one which is a multiple of the derivative of the denominator and the other a constant. In the present case, this is easy to do. The derivative of x 4x + is x 4, and clearly the numerator of our integral, x + 3, is equal to (x 4) + 7. So we write the integral as the sum of two integrals: x 4 7 x + 3 dx = dx + dx x 4x + x 4x + x 4x + (4) The first integral on the right-hand side of Equation 4 can be found by means of the substitution y = x 4x +, and since d y = (x 4) d x, the integral becomes dy = log e y + C = log e (x 4x + ) + C. y The second integral on the right-hand side of Equation 4 is of the sort you learnt to evaluate in Subsection.. The trick of writing numerator = multiple of derivative of denominator + constant is known as splitting the numerator. It involves only very simple algebra, as the following example shows. COPYRIGHT 998 S57 V.

29 Example 5 x + dx. Split the numerator in the integral 3x + 4x + d (3x + 4x + ) = 6x + 4. So we write x + = a(6x + 4) + b. dx Equating the coefficients of x on both sides we find = 6a, i.e. a =/3, then equating the constant terms we find = 4a + b, so that b = /3. Solution4The derivative of the denominator is Thus 6x + 4 x + dx = dx dx 4 3x + 4x + 3 3x + 4x + 3 3x + 4x + 4 x dx. Split the numerator in the integral x x+3 COPYRIGHT 998 S57 V.

30 x The technique of splitting the numerator can also be used to find integrals such as dx where the 5 x x integrand has the square root of a quadratic function of x in the denominator, and a linear function of x in the numerator. To find this integral, we again write the numerator as a multiple of the derivative of the quadratic d (5 x x ) = x ], plus a constant; so that for this example function in the denominator [in this case dx derivative of we have quadratic constant } x = a( x) + b numerator 678 Equating the coefficient of x, and the constant terms, on each side of this equation, we find a = /, b =. x x So dx = dx + dx 5 x x 5 x x 5 x x The first integral on the right-hand side here can be found by the substitution y = 5 x x, so that dy = ( x)dx, and we have x dy = y + C = 5 x x + C dx = 5 x x y COPYRIGHT 998 S57 V.

31 The second integral is one of those that you learnt to evaluate in Subsection.. We complete the square in the denominator, to obtain dx dx = 5 x x 6 (x + ) and make the substitution y = x +, followed by the substitution y = x + dx = arcsin +C 6 5 x x So, finally, x dx = 5 x x 6 sin u to obtain x + 5 x x + arcsin +C 6 Question T6 Find the integral t = r dr, where a, b and c are constants, for the case a =, b = 4 and c = 3. ar + br c (Such integrals arise in physics when the motion of an object that is moving under an inverse square law of force is considered. Its distance r from the centre of force will be related to time t by such an integral.)4 COPYRIGHT 998 S57 V.

32 .4 Substitutions involving hyperbolic functions You should know how to evaluate integrals of the form: o dx by means of the substitution x = sinu x o dx x + by means of the substitution x = tanu o dx x using partial fractions. Looking at this list, it may occur to you that there are two integrals which are missing from it, although they are very similar to the integrals that do appear there: we have in mind the integrals dx 4and4 dx x + x In this subsection, we will show how to find these integrals (and many more which are related to them). COPYRIGHT 998 S57 V.

33 The integral dx is equal to arcsinx + C; and one way to prove this is to make the substitution x x = sinu. This substitution gets rid of the square root in the denominator, by virtue of the trigonometric identity cosu + sin u =; x becomes sin u = cos u, and this gives us a clue as to how to proceed with the integral dx. We recall that the hyperbolic functions cosh and sinh satisfy an identity very similar to + x the one satisfied by cos and sin, but with a minus sign present, i.e. instead of cos u + sinu =, we have coshu sinhu =. Thus we can get rid of the square root in + x if we make the substitution x = sinhu, d which gives + x = + sinh u = cosh u. Since (sinh u) = cosh u, we also have dx = coshu du. du So, with the substitution x = sinhu, the integral dx becomes very easy, and specifically we have + x cosh u du = u + C dx = + x cosh u If x = sinhu then u = arcsinhx. Thus finally we have the result COPYRIGHT 998 S57 V.

34 dx = arcsinh x + C x + We can also exploit the identity cosh u sinhu = to evaluate the integral rewritten in the form sinh u = (5) dx. The identity can be x cosh u, which suggests that we make the substitution x = coshu. What does the integral dx become if we make the substitution x = coshu? x If x = coshu then u = arccoshx, so that dx = arccosh x + C x COPYRIGHT 998 (6) S57 V.

35 Now that you know how to find the basic integrals dx and dx, you should not find it hard + x x to adapt the method to find integrals involving similar square roots. Here is an example. Example 6 Find the integral dx x Solution4The experience that we gained in deriving Equation 5 suggests that we need to make a substitution of the form x = asinhu, where a is a suitably chosen constant. We choose a so that the identity coshu sinh u = can be used to turn 4 + 9x into a multiple of coshu. If we substitute x = asinhu into 4 + 9x, it becomes 4 + 9a sinh u which is equal to coshu if we choose 9a = 4, i.e. a = /3. So the required substitution is x = (/3)sinhu. We then have dx = (/3)coshudu, and the integral becomes dx = ( 3) cosh u du = du = u + C 3 cosh u x COPYRIGHT 998 S57 V.

36 Since x = (/3)sinhu we have u = arcsinh(3x/) and so finally we obtain 3x dx = arcsinh + C x dx or dx where a and b are positive When you have to evaluate integrals of the form a + bx bx a constants, you should start by asking yourself What substitution will get rid of the square root in the denominator when I use the identity cosh u sinhu =? If you make this your goal, you will be able to decide whether you need to make a substitution of the form x = asinhu or of the form x = acoshu, and you will also be able to determine the required value of the constant a. Use this approach in answering the following question. Question T7 Find the integrals: (a) dx, (b) dx 4 8x + 3 4x COPYRIGHT 998 S57 V.

37 You can, of course, combine substitutions involving hyperbolic functions with the techniques of completing the square and splitting the numerator, in order to find even more integrals. For example, consider the integral I= dx. 4x + x Write the quadratic function 4x + x in completed square form. What substitution should we make in order to evaluate COPYRIGHT 998 S57 V. y 4 dy?

38 Question T8 r dr. Find the integral r + r (Hint: This question requires you to split the numerator; this will leave you with two integrals, one to be found by completing the square in the denominator and substituting a hyperbolic function.)4 We have seen that substitutions of hyperbolic functions enable us to find many integrals where the denominator of the integrand is the square root of a quadratic function. They can also be used to integrate other functions involving a square root of this sort, as in the following example. COPYRIGHT 998 S57 V.

39 Example 7 Find the integral x dx. Solution4We first try the substitution x = coshu. Then x = sinh u and dx = sinhu du. So our integral becomes sinh u du. This may not immediately seem like much of an improvement on the integral we started with. However, there is an identity involving hyperbolic functions which will help us here: the identity cosh(u) = + sinhu. Rearranging this gives us sinh u = [cosh (u) ], and substituting this into the integral sinh u du gives sinh u du = cosh (u) du du = 4 u sinh (u) + C We could put u = arccosh x in here We now need to express our answer in terms of x. Of course, we could simply put u = arccoshx everywhere in the answer, but it is in fact possible to simplify the expression sinh(arccoshx) if we recall another identity, sinh(u) = coshu sinhu, and use the result sinh u = COPYRIGHT 998 cosh u. S57 V.

40 Then we can write sinh (u) = cosh u cosh u = x x, and so the final answer is x dx = x x arccosh x + C4 Study comment The indefinite integral in Example 7 was quite hard, partly because of the work required to express the final answer in terms of x. In the following question you are asked to evaluate a similar definite integral; if you transform the limits of integration when you make the substitution, you will not need to obtain a final answer in terms of x. Question T9 The length of the section of the parabola y = x between the points (, ) and (, ) is given by the integral + 4x dx. Evaluate this integral, using the substitution x = sinh u and the identity cosh(u) = coshu.4 COPYRIGHT 998 S57 V.

41 We could carry on almost indefinitely, and work through many examples of integrals which can be found using the substitution of a hyperbolic function. (For example, the integral x 4x + dx can be found by the method described in Example 7, but you need to complete the square first.) However, we do not have the space; besides, it would become very tedious. You should simply bear in mind that whenever you encounter an integral that contains the square root of a quadratic function of x, a sinh or cosh substitution may well enable you to simplify it, if you cannot think of anything else. Of course, there may be quicker ways to do an integral of that sort! What is the quickest way to find the integral x COPYRIGHT 998 x dx? S57 V.

42 .5 Using trigonometric identities You should already know how to find integrals like cos x sin x dx, but we will explain how it is done for the sake of completeness. Since cosx is the derivative of sinx, the substitution y = sinx can be used to simplify the y dy = 3 y 3 + C = 3 sin 3 x + C. You may not yet have encountered similar integrals involving hyperbolic functions, such as cosh x sinh x dx, but the same sort integral; and with this substitution, the integral becomes of approach will work with those: coshx is the derivative of sinhx, so, with the substitution y = sinhx this integral similarly becomes cosh x sinh x dx = y dy = 3 y 3 + C = 3 sinh 3 x + C COPYRIGHT 998 S57 V.

43 How can we find the equally simple-looking integral cos3 x dx? If we use the trigonometric identity cos x + sin x =, we can write this integral as the sum of two integrals that can be found. We proceed as follows: cos 3 x dx = cos x cos x dx = cos x( sin x) dx 3 sin x = cos x dx cos x sin x dx = sin x 3 sin 3 x + C A similar approach will enable you to integrate any product of powers of cosx and sinx in which either cosx or sinx (or both) is raised to an odd power. Here is an example. COPYRIGHT 998 S57 V.

44 Example 8 Find the integral sin 6 x cos5 x dx. (sin 6 x cos 4 x) cos x dx. We then express cos 4 x in terms of sinx, writing cos4x = ( sin x), so the integral becomes sin 6 x( sin x) cos x dx. We now make the Solution4We write this integral as substitution y = sinx, dy = cosxdx, to obtain the integral y 6 ( y ) dy = (y 6 y8 + y ) dy = So finally, sin 6 x cos5 x dx = 7 9 y y + y +C sin x sin 9 x + sin x + C Question T Evaluate the definite integrals: (a) π π sin 3 x dx 4(b) sin 3 x cos x dx 4 COPYRIGHT 998 S57 V.

45 A very similar method can be used to find integrals of odd powers of coshx or sinhx; you simply need to employ the identity cosh x sinhx =. Find the integral sinh 3 x dx. You now know how to integrate odd powers of cosx and sinx (or coshx and sinhx); what about even powers? How can we find, for example, cos x dx and sin x dx? Here we can make use of the identity cos(x) = cos x = sin x which can be rearranged to give two very useful relations: cos x = [ + cos ( x)] (7a) sin x = [ cos ( x)] (7b) COPYRIGHT 998 S57 V.

46 cos x = [ + cos ( x)] (7a) sin x = [ cos ( x)] (7b) We can use Equation 7a to find cos x dx. We write cos x dx = [ + cos ( x)] dx = dx + cos ( x) dx The first of these integrals is equal to so cos x dx = x + C and the second is equal to x + sin ( x) + C 4 Use Equation 7b to find the integral sin x dx. Express cos(x/) in terms of cosx. COPYRIGHT 998 S57 V. 4 sin ( x) + C,

47 Find cos (x ) dx To evaluate higher even powers of cosx or sinx, we can simply use the identities in Equations 7a and 7b more than once, as in the following example. Example 9 Find the integral cos 4 x dx. Solution4Using Equation 7a, cos x = [ + cos ( x)] (Eqn 7a) we have cos 4 x dx = 4 [ + cos ( x)] dx = 4 dx + cos ( x) dx + 4 cos ( x) dx (8) We can easily integrate the first two terms here, and to deal with the last term we use Equation 7a again, with x in that equation replaced by x, giving cos ( x) = [ + cos (4x)]. Substituting this in Equation 8 gives us cos 4 x dx = 83 dx + cos ( x) dx + 8 cos (4x) dx COPYRIGHT 998 = S57 V. 3 8 x + 4 sin ( x) + 3 sin (4 x) + C 4

48 With other integrals involving even powers of cosx and sinx, it may be necessary to use the identity cosx + sin x =, as well as the identities in Equations 7a and 7b. For example, consider the integral cos x sin x dx. One way to evaluate this would be to replace sinx by cos x, thus turning the integral into cos x dx cos 4 x dx. These are both integrals that have been covered in this module. Another way to evaluate cos x sin x dx is to make use of the identity sin(x) = sinxcosx (9) (sinxcosx), Since the integrand is simply we can write the integral as You should also be familiar with this integral. 4 sin ( x) dx. Identities analogous to those in Equations 7a, 7b cos x = [ + cos ( x)] (Eqn 7a) sin x = [ cos ( x)] (Eqn 7b) and Equation 9 hold for hyperbolic functions. COPYRIGHT 998 S57 V.

49 cosh x = [ + cosh ( x)] () sinh x = [cosh ( x) ] () sinh(x) = sinhxcoshx () These may be used to evaluate integrals of even powers of coshx and sinhx. We have already used Equation to evaluate sinh x dx in Example 7. You can practise using these identities by trying the following question. Question T Find the integral sinh 4 x dx.4 Repeated use of the identities in Equations 7a, 7b and 9 to can, however, become rather tedious in working out integrals of high even powers of cosx and sinx (or coshx and sinhx). In the next subsection, we show you an alternative, less laborious, way of finding such integrals. COPYRIGHT 998 S57 V.

50 .6 Reduction formulae We have just shown you one way of finding the integral cos x dx. This integral can also be found using integration by parts, and the method is worth describing, as it will lead to an elegant means of finding integrals of high powers of cosx and sinx. We introduce the notation I = cos x dx. Applying the formula for integration by parts dg df f (x)g(x) dx = F(x)g(x) F(x) dx dx 4where4 dx and taking f(x) = g(x) = cosx, so that F(x) = sinx and = f (x) (3) dg = sin x, we find dx I = cos x) dx {x cos {x dx = sin {x cos {x sin {x ( 4sin4 3 f (x) g( x ) F( x ) g( x ) F( x ) dg dx I = sin x cos x + sin x dx = sin x cos x + [ cos x] dx = sin x cos x + dx cos x dx 443 the integral I Pay particular attention to the fact that the integral I has appeared on the right-hand side of the final expression. COPYRIGHT 998 S57 V.

51 If we replace dx by x + C we can write the result of the above calculation in the form I = sinxcosx + x + C I we can see that by rearranging this equation we have I = sinxcosx + x + C so that the required integral is given by I = cos x dx = sin x cos x + x+c (4) Let us see what would happen if we applied the same method to the integral of some other power of cosx. In fact, we will not specify the power; we will consider the general integral In = cos n x dx (where n is a positive integer greater than ). We write In as (cos x)(cos n ) dx and apply Equation 3, dg df f (x)g(x) dx = F(x)g(x) F(x) dx dx 4where4 dx taking f (x) = cosx and g(x) = cosn x. Then F(x) = sinx, and COPYRIGHT 998 S57 V. = f (x) (Eqn 3) dg = (n ) cos n x sin x. dx

52 So Equation 3 dg df f (x)g(x) dx = F(x)g(x) F(x) dx dx 4where4 dx = f (x) (Eqn 3) gives the result In = sin x cos n x + (n ) cos n x sin 3x dx replaced by cos x I = sin x cos n x + (n ) cos n x ( cos x) dx = sin x cos n x + (n ) cos n x dx (n ) cos n x dx 443 (5) this is In Again, the integral I n that we are interested in appears on the right-hand side; unfortunately so does the integral cos n x dx. This is not an integral that we can evaluate as it stands, but note that it is of the same form as the integral In that we are trying to find; we can call it In. With this notation, if we take all the terms in In to the left-hand side of Equation 5, it then becomes COPYRIGHT 998 S57 V.

53 nin = sinxcosn x + (n )In i.e. cos n x dx = n sin x cos n x + cos n x dx n n The reason why Equation 6 is useful is that if we have already evaluated (6) cos n x dx, it allows us to write cos n x dx very quickly. In this way, it is possible to build up a whole sequence of integrals of powers of cosx. For example, we have already evaluated cos x dx (see Equation 4); and we can use this to find cos 4 x dx. We put n = 4 in Equation 6, to obtain down cos 4 x dx = 3 sin x cos 3 x 3sin x cos x 3x sin x cos3 x + cos x dx = + + +C Now that we know cos 4 x dx, we can use it in Equation 6, setting n = 6, to find cos 6 x dx, and so on. COPYRIGHT 998 S57 V.

54 Substitute n = 3 in Equation 6, n cos n x dx = n sin x cos n x + n cos n x dx and hence find (Eqn 6) cos3 x dx. cos5 x dx (you may make use of Equation 7) cos x dx = 3 sin x cos x + 3 cos x dx Find = sin x cos x + cos x dx = sin x cos x + sin x + C COPYRIGHT 998 S57 V. (Eqn 7)

55 Formulae such as Equation 6, n cos n x dx = n sin x cos n x + n cos n x dx (Eqn 6) which relate an integral involving a power of some function (in the above case In) to a similar integral involving a lower power of the same function (in the above case In ), are known as reduction formulae. They enable us to build up a whole sequence of integrals, starting from an integral that is easy to evaluate. There is no need to memorize reduction formulae, but you should be aware that they exist, and be able to look them up and use them. We will present you here with two more examples of reduction formulae. As you might expect, there also exists a reduction formula that enables us to find integrals of powers of sinx. Here it is: sin n x dx = n cos x sin n x + n sin n x dx n (8) The derivation of this formula is very similar to the derivation of Equation 6, and we leave it for you to do in the following question. COPYRIGHT 998 S57 V.

56 Question T (a) Derive Equation 8. n sin n x dx = n cos x sin n x + n sin n x dx (Eqn 8) (Hint: Start by integrating In = sin n x dx by parts, taking f(x) = sinx and g(x) = sinn x in Equation 3.) dg df f (x)g(x) dx = F(x)g(x) F(x) dx dx 4where4 dx (b) Use Equation 8 to find sin 3 x dx.4 COPYRIGHT 998 S57 V. = f (x) (Eqn 3)

57 A third useful reduction formula deals with integrals of the form x n e ax dx where a is a positive constant and n is a positive integer. Again we use integration by parts to derive the reduction formula. In the integral, we take dg f(x ) = e a x and g (x) = xn, then F(x) = e ax and = nx n and substituting these expressions into a dx Equation 3, dg df (Eqn 3) f (x)g(x) dx = F(x)g(x) F(x) dx dx 4where4 dx = f (x) e ax x n nx n e ax dx ax n we obtain e x dx = {{ a a f ( x ) g( x ) 443 { g( x ) ( ) F( x ) so that dg dx F( x ) n x n e ax dx = x n a e ax + a x n e ax dx Equation 9 shows that we can express power of x. x n e ax dx COPYRIGHT 998 in terms of S57 V. (9) x n e ax dx, an integral involving one less

58 In the simplest case, if n = in Equation 9, n x n e ax dx = x n a e ax + a x n e ax dx we can find x e ax dx, and so obtain x x (Eqn 9) x e ax dx = a e ax + a e ax dx = a e ax a e ax + C Given that xe ax dx = x ax ax e e + C, find a a COPYRIGHT 998 x e ax dx. S57 V.

59 .7 Mixed examples Aside You may be aware of the existence of algebraic computing programs such as: Mathematica, Reduce, Maple, Mathcad or Derive, which can find indefinite integrals, and evaluate definite integrals as functions of the parameters appearing in them. If so, you may be wondering why you should bother to learn advanced integration techniques; why not just key in the integral and let the program do the work? Such programs have their limitations. They may not give you the answer in the form you are expecting, for instance, in Example 6 we found 3x dx = arcsinh + C x but using Derive you would obtain dx = log e (9x + 4) + 3x + C x It is not immediately obvious that these two answers are the same. In the case of improper but convergent integrals, the program may decide (incorrectly) that the integral cannot be evaluated, and you may have to make a substitution in order to make the integral acceptable to the program. Such programs can be very useful, and they certainly alleviate the tedious business of calculating unpleasant integrals, but you would be very unwise to use them without understanding the basic principles of integration. COPYRIGHT 998 S57 V.

60 How to tackle a general integral You have learnt several tricks that will enable you to find a wide variety of integrals. We do not intend to summarize them here. First, their uses are so widespread that such a summary would be lengthy and boring; second (and more important) we do not want you to feel that you must learn by heart a long list of different types of integral and the methods that will work for them. Certainly you should have these methods as part of your mathematical furniture, but you should think of them as techniques to be applied in a trial and error way. It is not a disaster if the first method you try does not work; you simply have to try something else. You should bear in mind too, that many integrals can only be found by a combination of the techniques you have learnt here; it may, for example, be necessary for you to use more than one substitution, and perhaps combine substitutions with algebraic manipulation or use of identities. As you gain more experience with integration, you will begin to see automatically what sort of approaches are likely to prove productive with a given integral. The following two questions should serve as revision of the integration techniques we have discussed. COPYRIGHT 998 S57 V.

61 Question T3 Find the following integrals: dx (Hint: Use partial fractions.) (a) x x (b) cos 7 ( x)sin 4 ( x) dx (Hint: Use trigonometric identities.) 4x dx (Hint: Split the numerator; complete the square in the denominator; use a hyperbolic (c) + 4x + x function substitution.) (d) (x )3 dx (Hint: Substitute a hyperbolic function; use the answer to Question T.)4 COPYRIGHT 998 S57 V.

62 Question T4 Find the following integrals: 3x + dx,4(b) (a) (x + x + )(x ) π/ sin 6 x dx,4(c) COPYRIGHT 998 x dx.4 x 4 S57 V.

63 3 Some definite integrals 3. Integrals with an infinite upper or lower limit Definite integrals in which one or both of the limits is infinite occur very often in physics. Here are some examples: V= Q 4πε x dx () () () (3) r dx, where x and a are constants, and a > (x x ) + a m v = 4π πkt 3/ v3exp ( mv kt ) dv x n e ax dx, where n is a non-negative integer, and a > COPYRIGHT 998 S57 V.

64 Definite integrals over an infinite range of integration are known as improper integrals. The way to evaluate them is to replace the infinite limit by a very large (but finite) number, evaluate the integral in the usual way, and then see what happens to your result as the large number becomes larger still. More formally, we think of an improper integral as a limit: b a a lim f (x) dx f (x) dx = b b and f (x) dx = lim a b f (x) dx If you can find the indefinite integral F(x) + C = f (x) dx a f (x) dx, evaluating an improper definite integral of the form simply requires you to know how the function F(x) behaves when x becomes very large. Often this is a obvious. COPYRIGHT 998 S57 V.

65 For example, to evaluate the integral in Equation, V= Q 4πε x dx (Eqn ) r we think of it as the limit as R of R R Q dx = Q = Q + Q 4πε x 4πε R 4πε r 4πε x r r As R becomes very large, /R becomes smaller and smaller, so that as R the first term tends to zero. So we are left with Q Q V= dx = 4πε r 4πε x r COPYRIGHT 998 S57 V.

66 Sometimes it may not be immediately obvious what happens to F(x) as x becomes very large. For example, at the start of Subsection. we showed that dx = log e (x + ) log e (x + ) + C (x + )(x + ) dx. Suppose that we want to find the definite integral (x + )(x + ) lim [ log e (x + ) log e (x + )] a ; a We can express this integral as what happens to the difference between two logarithms when their arguments, i.e. (x +) and (x + ) both become very large? We can see what happens if we write this difference of logarithms as the logarithm of a fraction: x + log e (x + ) log e (x + ) = log e x + As x becomes very large, the fraction tends towards the value, and its logarithm tends to zero. dx = log e =.693 So (x + )(x + ) COPYRIGHT 998 S57 V.

67 What is the limit as x of 3x + x + 5? x 4x Another potentially tricky situation arises when we consider the integral in Equation 3, namely x n e ax dx. We are already part of the way in evaluating the indefinite integral for it in Subsection.6: n x n e ax dx = x n a e ax + a x n e ax dx xn e ax dx ; we derived a reduction formula (Eqn 9) We can easily convert Equation 9 into a reduction formula involving definite integrals between and, by putting in the limits: x n e ax dx = a [ x n e ax ] + n n ax x e dx a COPYRIGHT 998 (4) S57 V.

68 We now need to know what happens to the function x n e ax as x becomes very large. There is something of a dilemma here: e ax becomes very small for large x (remember that a is positive), but xn becomes very large; which one wins? Although we will not prove it here, the rule is as follows: xne ax as x (5) for any n and any positive constant a Using this rule to evaluate the term [ x n e ax ] in Equation 4, n ax n n ax (Eqn 4) [ x e ] + a x e dx a we see that the function is zero at the upper limit; it is also zero at the lower limit (because of the factor xn). So Equation 4 becomes x n e ax dx = x n e ax dx = n n ax x e dx a COPYRIGHT 998 (6) S57 V.

69 Evaluate the integral e ax dx Evaluate the integrals x e ax dx and n x e ax dx, using Equation 6. x n e ax dx = a x n e ax dx (Eqn 6) COPYRIGHT 998 S57 V.

70 You may be able to see a pattern emerging. If we now use Equation 6 n x n e ax dx = a x n e ax dx (Eqn 6) to work out x 3 e ax dx, we find x 3 e ax dx = 3, a4 x 4 e ax dx = 4 3 4and so on. a5 The general result is x n e ax dx = n! a n + (7) Evaluate the integral x 7 e x dx. COPYRIGHT 998 S57 V.

71 So far, we have considered improper integrals which could be evaluated without using a substitution. Of course, if you want to evaluate an improper integral using a substitution, you will need to transform the infinite limit of integration appropriately. Sometimes it will turn into a finite quantity, as in the next example. Example Evaluate the integral dx appearing in Equation. (x x ) + a Solution4This integral can be found using the substitution x x = atanu. Then dx = asec udu, and (x x ) + a = a secu. When x is very large and positive, so is x x = atanu. If atanu is very large, then u is close to π/, and as tanu tends to infinity, u tends to π/; so the upper limit of integration becomes π/. Similarly, the lower limit of integration becomes π/. Thus π/ π dx = du = 4 a a π/ (x x ) + a COPYRIGHT 998 S57 V.

72 It is usually quite straightforward to change infinite limits of integration on making a substitution of the form u = g(x) or x = h(u); you need only ask yourself What does u tend to as x gets very large? Of course, often you will find that the infinite limit of integration is still infinite. This is the case in the following question. Question T5 Using the substitution u = v, and Equation 7, x n e ax dx = n! a n + (Eqn 7) evaluate the integral v3 exp ( mv kt ) dv that appears in Equation.4 o COPYRIGHT 998 S57 V.

73 3. Gaussian integrals Integrals of the form x r exp ( ax ) dx, where r is a positive integer or zero, and a is a positive constant, are particularly common in physics. You have already seen one example of this sort in Equation m v = 4π πkt 3/ v3exp ( mv kt ) dv (Eqn ) (and evaluated it in Question T5). Many others like this arise in the kinetic theory of gases, in quantum mechanics, and also in probability theory (which you may find yourself using to analyse experimental data). In the case where r is an odd integer, such integrals can easily be evaluated, using the same substitution that you employed in Question T5. To make it clear that r is odd, we will write r = n +, where n is any positive integer or zero, and consider the integral x n + exp ( ax ) dx which we write as x n exp ( ax )x dx. COPYRIGHT 998 S57 V.

74 We now make the substitution y = x. Then x dx = dy, and xn = (x)n = yn. So x n + exp ( ax ) dx = y n e ay dy We know how to evaluate y n e ay dy ; from Equation 7, n! x n e ax dx = a n + it is equal to (Eqn 7) n!. Thus a n + n! x n + exp ( ax ) dx = a n + (8) COPYRIGHT 998 S57 V.

75 Evaluate x 5 exp ( 3x ) dx. Notice, too, that because we can relate x n + exp ( ax ) dx to y n e ay dy by the substitution y = x, and we can find the latter integral (for any given value of n) by applying Equation 9 n (Eqn 9) x n e ax dx = x n a e ax + a x n e ax dx as many times as is necessary, there is no problem in finding the indefinite integral we want to do so. COPYRIGHT 998 S57 V. x n + exp ( ax ) dx, should

76 The situation is very different for the indefinite integral x r exp ( ax ) dx where r is an even integer. If you try x the substitution = y, you will quickly find that it does not help at all there is an awkward factor of y left over in the integrand. In fact, such indefinite integrals cannot be evaluated in terms of familiar functions (such as exp, loge, powers of x and so forth). The simplest of these integrals (with r =, a = ) is exp ( x ) dx. This integral is in fact used to define a new function of x, known as the error function, erf(x); the precise definition is erf (x) = x exp ( y ) dy π It is possible to evaluate this integral for different values of x by numerical techniques, and the results have been tabulated and can be looked up in books (indeed, some calculators have an erf button). If you study advanced probability theory in future, you are bound to come across this function. However, here we will be concerned just with the definite integral exp ( x ) dx, and definite integrals such as x r exp ( ax ) dx (where r is an even integer) which can be related to it. Such integrals are called Gaussian integrals. COPYRIGHT 998 S57 V.

77 It is possible to evaluate exp ( x ) dx exactly by methods that are more advanced than those discussed in FLAP; we will present the answer here, and then show that a whole host of other Gaussian integrals can be found in terms of this integral. The basic result is exp ( x ) dx = π (9) We can use Equation 9 to find the integral exp ( ax ) dx. We simply make the substitution dy, and the limits of integration are still and. So a and from Equation 9, Then dx = exp ( ax ) dx = π a COPYRIGHT 998 exp ( ax ) dx = (3) S57 V. y= a x. exp ( y ) dy, a

78 Evaluate exp ( ax ) dx. We can now use integration by parts to derive a reduction formula relating x r exp ( ax ) dx to x r exp ( ax ) dx. To make it clear that r is an even integer, we write r = n, where n is any positive integer. We can write the integral x n exp ( ax ) dx as x exp ( ax ) x n dx, and apply Equation 3, dg df f (x)g(x) dx = F(x)g(x) F(x) dx dx 4where4 dx taking f (x) = x exp ( ax ) 4and4 g(x) = x n COPYRIGHT 998 S57 V. = f (x) (Eqn 3)

79 Then F(x) = dg exp ( ax ) (as you can easily check by differentiating), and = (n )x n. a dx So n x n exp ( ax ) dx = a x n exp ( ax ) + a x n exp ( ax ) dx (3) Now from the rule in Equation 5, xne ax as x (Eqn 5) xn e ax (for any n and any positive constant a), we know that tends to zero as x tends to infinity; and since exp( ax) is smaller than e ax when x is large, we can be sure that xn exp( ax) also tends to zero as x tends to infinity. Since n, xn exp( ax) = when x =. So the first term on the right-hand side of Equation 3 is zero, leaving us with the reduction formula x n exp ( ax ) dx = n n x exp ( ax ) dx a COPYRIGHT 998 S57 V. (33)

80 We can now use this reduction formula, n x exp ( ax ) dx = n n x exp ( ax ) dx a (Eqn 33) π a (Eqn 3) and Equation 3, exp ( ax ) dx = to work out values of x n exp ( ax ) dx for successively higher values of n. Evaluate x exp ( ax ) dx. Gaussian integrals may sometimes appear in a disguised form. In the following question you have to make a substitution before it becomes clear that you are dealing with a Gaussian integral. COPYRIGHT 998 S57 V.

81 Question T6 Evaluate the integral x 3/ e 5 x dx. (Start by making the substitution x = y.)4 Finally, integrals of the form Equation 3, exp ( ax + bx) dx can be easily related to the integral appearing in exp ( ax ) dx, π (Eqn 3) a by the trick of completing the square in the exponent. The following example shows you how to proceed. exp ( ax ) dx = exp ( ax ) dx = COPYRIGHT 998 S57 V.

82 Example Evaluate the integral exp ( x + 4x) dx Solution4We write x + 4x in completed square form: x + 4x = (x x) = [(x ) ] = (x ) + Thus the integral can be written as exp[ (x ) ]e dx. We now make the substitution y = x, dy = dx. The limits of integration are still and ; so, taking the constant e outside the integral sign, we have exp[ (x ) ]e dx = e exp ( y ) dy COPYRIGHT 998 S57 V.

83 Substituting a = in Equation 3, exp ( ax ) dx = exp ( ax ) dx = we find that exp ( y ) dx = exp ( x + 4x) dx = e π a (Eqn 3) π ; so π Question T7 Evaluate the integral exp ( x 3x) dx 4 COPYRIGHT 998 S57 V.

84 3.3 Line integrals This subsection discusses a type of definite integral that is known as a line integral. Line integrals can be used (among other applications) to evaluate the work done by a force, and we will introduce them in that context. Suppose that an object moves along the x-axis under the influence of a constant force F x in the x-direction. The work done by the force in moving the object from x = a to x = b is W = sxfx, where sx is the displacement of the object. If the force is not constant, but instead varies with x, Fx(x) say, then you may already know that the b work done in moving the object from x = a to x = b is given by the definite integral Fx (x) dx. a To derive this result, we divide the interval a x b into many small subintervals: we introduce n + values of x such that a = x < x < < x n < x n+ = b, and we define xi = x i+ x i, where i is any integer in the range i n and let x be the largest of these subintervals. We can then say that over any one of these small intervals, the force is approximately constant, so that the work done by the force in moving the object from xi to xi+ is approximately equal to Fx(xi) xi. Then the total work done by the force is approximately given by the sum n Fx (xi ) xi (34) i = COPYRIGHT 998 S57 V.